What is the integral of 1/(sqrt(4+x-x^2)) ?

The integral of 1/(sqrt(4+x-x^2)) is arcsin(1/(sqrt(17))(2x-1))+C

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integral of 1/(sqrt(4+x-x^2))

\int \frac{1}{4 + x - x ^{2}} d x

arcsin (\frac{1}{17} (2 x - 1)) + C

Solution steps

\int \frac{1}{4 + x - x ^{2}} d x

Complete the square

4 + x - x^{2} : - (x - \frac{1}{2})^{2} + \frac{17}{4}

= \int \frac{1}{- ( x - \frac{1}{2} ) ^{2} + \frac{17}{4}} d x

Apply u-substitution

: \int \frac{2}{- 4 u ^{2} + 17} d u

= \int \frac{2}{- 4 u ^{2} + 17} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int \frac{1}{- 4 u ^{2} + 17} d u

Apply Trigonometric Substitution

: \int \frac{1}{2} d v

= 2 \cdot \int \frac{1}{2} d v

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \frac{1}{2} \cdot \int 1 d v

Integral of a constant:

\int a d x = a x

= 2 \cdot \frac{1}{2} \cdot 1 \cdot v

Substitute back

= 2 \cdot \frac{1}{2} \cdot 1 \cdot arcsin (\frac{2}{17} (x - \frac{1}{2}))

Simplify

2 \cdot \frac{1}{2} \cdot 1 \cdot arcsin (\frac{2}{17} (x - \frac{1}{2})) : arcsin (\frac{1}{17} (2 x - 1))

= arcsin (\frac{1}{17} (2 x - 1))

Add a constant to the solution

= arcsin (\frac{1}{17} (2 x - 1)) + C

\frac{d}{d x} (4 x^{2} + x y + 2 y^{2})

\frac{\partial}{\partial x} (x^{2} + y^{2} + z^{2})

\int \frac{1}{13 x - 2 x ^{2} - 15} d x

\frac{d}{d x} ((3 x^{5}) (6 x^{4} + x)^{4})

e^{x^{2}}

What is the integral of 1/(sqrt(4+x-x^2)) ?
The integral of 1/(sqrt(4+x-x^2)) is arcsin(1/(sqrt(17))(2x-1))+C